Answer:
115.891
Explanation:
Molarity is equal to moles/liters so to get moles we multiply molarity by liters, after this we multiply the moles by the total atomic weight of the compound to get grams
Answer:
Explanation:
According to Bronsted-Lowry acids or base theory , the reagent capable of giving hydrogen ion or proton will be acid and that which accepts hydrogen ion or proton will be base .
C₉H₇N + HNO₂ ⇄ C₉H₇NH⁺ + NO₂⁻
If K > 1 , reaction is proceeding from left to right .
Hence HNO₂ is giving H⁺ or proton and C₉H₇N is accepting proton to form
C₉H₇NH⁺ .
Hence HNO₂ is bronsted acid and C₉H₇N is bronsted base .
B )
when K < 1 , reaction above proceeds from right to left . That means
C₉H₇NH⁺ is giving H⁺ so it is a bronsted acid and NO₂⁻ is accepting H⁺ so it is a bronsted base .
Hence , NO₂⁻ is a bronsted base and C₉H₇NH⁺ is a bronsted acid .
I think the answer you are looking for is D.
Hope this helps!!! :D
Answer:
(a) adding 0.050 mol of HCl
Explanation:
A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.
In the buffer:
1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>
1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>
-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-
Thus:
<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.
(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.
(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>
